Surface Area, Volume & Compound Solids

Mathematics· 5.4, 5.5· 25 min read

1. Surface Area & Volume of Standard Solids (Core)★★☆☆☆⏱ 8 min

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All standard solid formulae are provided in your exam paper, but you should be comfortable substituting values correctly and distinguishing between total surface area (all faces) and curved surface area (only curved faces, excluding flat ends for cylinders/cones).

📘 Definition

Standard Solid

A regular 3D shape with well-documented mensuration formulae, including cuboids, prisms, cylinders, spheres, pyramids and cones.

📐 Worked Example

Calculate the total surface area of a closed cylinder with radius 3 cm and height 8 cm. Leave your answer in terms of π.

1
Recall the formula for total surface area of a closed cylinder: $S A = 2 π r^{2} + 2 π r h$
2
$r = 3, h = 8, substitute: 2\pi (3)^2 + 2\pi (3)(8)$
3
$= 18\pi + 48\pi$
4
Total surface area = $66 π$ cm²

Exam tip:

Always check if the solid is open or closed (e.g., an open cylinder has no top circular face, so exclude 1 $π r^{2}$ term from total surface area).

2. Compound Solids (Core + Extended)★★★☆☆⏱ 7 min

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Compound solids are formed by joining two or more standard solids, or removing part of a standard solid. To calculate volume, add or subtract the volumes of the component parts as appropriate. For surface area, only count the external exposed faces: do not include faces that are glued together or removed.

📐 Worked Example

A compound solid is made by attaching a hemisphere of radius 4 cm to the flat circular top of a closed cylinder of radius 4 cm and height 10 cm. Calculate the total volume of the solid, use $π = 3.14$ .

1
Calculate volume of cylinder: $V_{cy l in d er} = π r^{2} h = 3.14 \times 4^{2} \times 10 = 502.4$ cm³
2
Calculate volume of hemisphere: $V_{h e mi s p h er e} = \frac{2}{3} π r^{3} = \frac{2}{3} \times 3.14 \times 4^{3} \approx 133.97$ cm³
3
Add the two volumes: Total $V = 502.4 + 133.97 = 636.37$ cm³

✓ Quick check

When calculating the surface area of the above compound solid, why do you not include the flat face of the hemisphere?
Reveal answer
That face is attached to the top of the cylinder, so it is not exposed externally.

3. Partial Solids (Core + Extended)★★★☆☆⏱ 5 min

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Partial solids are standard solids cut along a plane, e.g., half a sphere (hemisphere), half a cylinder cut lengthwise. For volume, take the fraction of the full solid's volume. For surface area, add the area of the new flat face created by the cut to the fraction of the original curved surface area.

📐 Worked Example

A solid wooden cylinder of radius 5 cm and height 12 cm is cut exactly in half along its vertical axis. Calculate the total surface area of one half of the cylinder, leave your answer in terms of π.

1
Curved surface area of half cylinder: $\frac{1}{2} \times 2 π r h = π r h = 60 π$ cm²
2
Area of the two half circular ends: equal to area of one full circle: $π r^{2} = 25 π$ cm²
3
Area of the new rectangular flat face: length = 12 cm, width = diameter = 10 cm, so area = 120 cm²
4
Total SA = $60 π + 25 π + 120 = 85 π + 120$ cm²

4. Extended Only: Frusta★★★★☆Extended only⏱ 7 min

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A frustum is formed when the top of a cone or pyramid is cut off with a plane parallel to its base. The removed top is a similar smaller version of the original solid. You can use similarity ratios to calculate the dimensions of the removed top, then subtract its volume/surface area from the original solid to find the values for the frustum.

📘 Definition

Frustum

The remaining part of a cone or pyramid after a similar smaller section is removed by a cut parallel to the base.

📐 Worked Example

A cone of height 20 cm and base radius 5 cm has a smaller cone of height 8 cm removed from its top, cut parallel to the base. Calculate the volume of the resulting frustum, leave your answer in terms of π.

1
Use similarity ratio: ratio of heights of small cone to original cone is 8:20 = 2:5, so radius of small cone is $(\frac{2}{5}) \times 5 = 2$ cm
2
Volume of original cone: $\frac{1}{3} π r^{2} h = \frac{1}{3} π \times 25 \times 20 = \frac{500}{3} π$ cm³
3
Volume of small removed cone: $\frac{1}{3} π \times 4 \times 8 = \frac{32}{3} π$ cm³
4
Volume of frustum = $\frac{500}{3} π - \frac{32}{3} π = 156 π$ cm³

Exam tip:

For surface area of a frustum, do not forget to add the area of the top circular face of the frustum, and exclude the curved surface area of the removed small cone.

5. Common Pitfalls

Wrong move:

Adding internal joined faces when calculating surface area of compound solids

Why:

Faces glued between two solids are not exposed, so counting them overestimates the surface area

Correct move:

Only sum the area of all external exposed faces of the compound shape

Wrong move:

Confusing curved surface area and total surface area for cylinders, cones, and spheres

Why:

Questions often specify which value to calculate, using the wrong formula leads to lost marks

Correct move:

Circle the keyword 'curved' or 'total' in the question before selecting your formula

Wrong move:

Forgetting to add the area of the cut face when calculating surface area of partial solids

Why:

Cutting a solid creates a new flat exposed face that is not part of the original solid's surface area

Correct move:

Add the area of the new flat face to the fraction of the original surface area for partial solids

Wrong move:

Using the original cone's slant height for the frustum's curved surface area (Extended only)

Why:

The slant height of the frustum is the difference between the original cone's slant height and the removed small cone's slant height

Correct move:

Calculate slant heights for both original and removed solids using Pythagoras' theorem, then subtract to get the frustum's slant height

Wrong move:

Rounding intermediate values too early in multi-step calculations

Why:

Early rounding introduces errors that can make your final answer fall outside the accepted mark range

Correct move:

Keep all intermediate values in exact form (in terms of π, or stored in calculator memory) until the final step

6. Quick Reference Cheatsheet

Solid Type	Volume Formula	Surface Area Formula	Given in Exam?
Cuboid	$V = l w h$	$S A = 2 (l w + l h + w h)$	No (memorize)
Prism	$V = Area of cross-section \times length$	$S A = 2 \times cross-section area + perimeter of cross-section \times length$	No (memorize rule)
Cylinder	$V = π r^{2} h$	Curved: $2 π r h$ , Total: $2 π r^{2} + 2 π r h$	Yes
Sphere	$V = \frac{4}{3} π r^{3}$	$S A = 4 π r^{2}$	Yes
Cone	$V = \frac{1}{3} π r^{2} h$	Curved: $π r l$ , Total: $π r^{2} + π r l$ ( $l$ = slant height)	Yes
Pyramid	$V = \frac{1}{3} \times base area \times vertical height$	$S A = base area + sum of triangular face areas$	Yes
Frustum (Extended only)	$V = V_{or i g ina l} - V_{r e m o v e d}$	$S A = curved area of original - curved area of removed + area of top face$	No (calculate via similarity)
Compound Solid	$V = sum of component volumes (subtract removed parts)$	$S A = sum of external exposed faces only$	No

7. Frequently Asked

Do I need to memorize all mensuration formulae for the exam?

Most formulae are printed on the front of your exam paper, but you should memorize the volume and surface area formulae for cuboids and the volume rule for prisms, as these are not always explicitly listed.

Can I leave my answer in terms of π?

Yes, unless the question explicitly asks for a numerical value rounded to a specific number of significant figures or decimal places.

Going deeper

formula_sheetCIE IGCSE Maths 0580 Official SyllabusRefer to pages 22-23 for official mensuration formulae

What's Next

Now that you have mastered surface area and volume calculations for standard, compound, and partial solids (plus frusta for Extended tier), you are ready to tackle structured exam questions on mensuration, including real-world application problems like calculating container capacity or material required to manufacture 3D objects. Practice with past paper questions to identify common question patterns and traps, and ensure you can substitute values into given formulae quickly and accurately. Make sure you can switch between exact (in terms of π) and rounded answers as required by the question.